Show:limn→∞n√n!n=1e
So I can expand the numerator by geometric mean. Letting Cn=(ln(a1)+...+ln(an))/n. Let the numerator be called an and the denominator be bn Is there a way to use this statement so that I could force the original sequence into the form of 1/(1+1n)n
Answer
I would like to use the following lemma:
If limn→∞an=a and an>0 for all n, then we have
limn→∞n√a1a2⋯an=a
Let an=(1+1n)n, then an>0 for all n and limn→∞an=e. Applying (∗) we have
e=limn→∞n√a1a2⋯an=limn→∞n√(21)1(32)2⋯(n+1n)n=limn→∞n√(n+1)nn!=limn→∞n+1n√n!=limn→∞nn√n!+limn→∞1n√n!=limn→∞nn√n!
where we use (1) in the last equality to show that
limn→∞1n√n!=0.
It follows from (2) that
limn→∞n√n!n=1e.
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